Calculating Probability of Score Comparison between Two Distributions

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To determine the probability of a single value from distribution A being greater than a single value from distribution B, one can use the difference of the two distributions, C = A - B. By calculating the distribution function for C based on the known means and standard deviations of A and B, the probability P(C > 0) can be computed. In the provided example, Team A has a mean of 5 runs and Team B has a mean of 4 runs, with their respective standard deviations being 1 and 2 runs. This method allows for a straightforward calculation of the probability that Team A outscored Team B on any given day. Understanding this approach is essential for comparing scores across different distributions effectively.
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Forgive me if this is a silly question, but here goes:

Say you have two unrelated distributions (A and B) with known means and standard deviations. How would you determine the probability of any single value of A being greater than any single value of B?

The easiest example I can come up with for clarification would be something like this:

Over 1000 games baseball team A scores: μA=5 runs, σA=1 runs per game.
Over 1000 games baseball team B scores: μB=4 runs, σB=2 runs per game.

What is the probability on any given day that Team A outscored Team B? Sorry if this example seems elementary, but I am far from a statistician. Thanks for the help.
 
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Direct method: Let C = A-B (there are standard methods to get the distribution function for C, given the distribution functions for A and B). Then compute P(C > 0).
 
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